# Real Analysis

Open and Closed Sets

# Limit Points

Say $$p \in X$$ is a limit point of $$E \subset X$$ if every neighborhood of $$p$$ (i.e. $$N_r(p)$$ for all $$r$$) contains a point $$q \neq p$$ such that $$q \in E$$.

So, a point $$p$$ is not a limit point of $$E$$ if there exists a neighborhood $$N$$ such that $$N$$ does not contain any other points of $$E$$.

For example,

1. In $$\mathbb{R}$$, $$G = \{\frac{1}{n} : n \in \mathbb{Z}\}$$. $$0$$ is a limit point of $$G$$. In $$\mathbb{R}$$, a neighborhood of a point $$p$$ is an open interval $$(p - \epsilon, p + \epsilon)$$.

# Isolated Points

A point $$p \in X$$ is an isolated point of $$E \subset X$$ if $$p \in E$$ and $$p$$ is not a limit point of $$E$$.

For exmple,

1. All points of $$G = \{\frac{1}{n} : n \in \mathbb{Z}\}$$ are isolated.

# Interior Points

A point $$p \in X$$ is an interior point of $$E \subset X$$ if $$p \in E$$ and there exists a neighborhood $$N$$ of $$p$$ such that $$N \subset E$$.

For example,

1. $$G = \{\frac{1}{n} : n \in \mathbb{Z}\}$$ has no interior points.
2. In $$\mathbb{R}$$, $$\emptyset$$ has no limit points, no interior points, and no isolated points, for $$\mathbb{R}$$ has no isolated points but all points in $$\mathbb{R}$$ are interior and are limit points, and $$\mathbb{Q}$$ has no isolated points and no interior points but all points in $$\mathbb{Q}$$ are limit points.
3. In ($$\mathbb{R}$$, $$\text{discrete}$$), $$\emptyset$$ has no limit points, no interior points, and no isolated points, $$\mathbb{R}$$ has no limit points but all points in $$\mathbb{R}$$ are interior and isolated, and $$\mathbb{Q}$$ has no limit points but all points in $$\mathbb{Q}$$ are interior and isolated.

Theorem. If $$p$$ is a limit point of $$E$$, then every neighborhood of $$p$$ contains infinitely many points of $$E$$.

Proof (by contradiction). Assume there exists a neighborhood $$N$$ of $$p$$ with only finitely many points in $$E$$, call them $$e_1, e_2, \dots, e_n$$.

Let $$r = \min_{i = 1}^{n} \{d(p, e_i)\}$$ ($$r$$ exists because $$e_i$$ is finite). But $$N_r(p)$$ has no points other than $$p$$, a contradiction! QED.

# Open Sets

A set $$E$$ is open if every point of $$E$$ is an interior point of $$E$$.

For example,

1. In $$\mathbb{R}$$, any open interval $$(a, b)$$ is open.
2. In $$\mathbb{R}$$, $$\emptyset$$ is open.
3. In $$\mathbb{R}$$, $$\mathbb{R}$$ is open.

Lemma. Neighborhoods are open.

Proof. Consider a neighborhood $$N$$ around a point $$p$$ with radius $$r$$. Consider a point $$q \in N$$. So $$d(p, q) < r$$. Let $$a = d(p, q)$$. Let $$r' = r - a > 0$$. Consider $$N_{r'}(q)$$. For all points $$x \in N'$$, we have $$d(q, x) < r'$$. Then $$d(x, p) \leq d(x, q) + d(q, p) < r' + a = r.$$ Thus, $$x \in N$$. Therefore, $$N_{r'} \subset N_r$$. QED.

# Closed Sets

A set $$E$$ is closed if $$E$$ contains all its limit points.

For example,

1. In $$\mathbb{R}$$, a single point $$p$$ is closed.
2. In $$\mathbb{R}$$, $$\emptyset$$ is closed.
3. In $$\mathbb{R}$$, $$\mathbb{R}$$ is closed.
4. In $$\mathbb{R}$$, any closed interval $$[a, b]$$ is closed.
5. In $$\mathbb{R}$$, any half-open interval $$[a, b)$$ or $$(a, b]$$ is neither open nor closed.

Sets which are both open and closed is called clopen.

# Closing a Non-Closed Set

The closure of $$E$$, denoted $$\overline{E}$$, is the union of $$E$$ and the set of its limit points, denoted $$E'$$. In other words, $$\overline{E} = E \cup E'$$.

Theorem. $$\overline{A}$$ is a closed set.

Proof. To prove $$\overline{A}$$ is closed, we need to prove all limit points $$p$$ of $$\overline{A}$$ is in $$\overline{A}$$. In order to prove that, we show that for all limit points $$p$$ of $$\overline{A}$$, $$p$$ is a limit point of $$A$$, i.e. $$p \in A' \subset \overline{A}$$.

For all $$r > 0$$, consider the neighborhood $$N$$ around $$p$$ with radius $$r$$, i.e. consider $$N_r(p)$$. We need to show that there exists a point $$q \in N \cap A$$. Since $$p$$ is a limit point of $$\overline{A}$$, there exists a point $$t \in N \cap \overline{A}$$. Thus, $$t$$ is in $$A$$ or $$t$$ is a limit point of $$A$$. When $$t \in A$$, we have $$q = t \in A$$. When $$t$$ is a limit point of $$A$$, consider a neighborhood $$M$$ around $$t$$ such that $$M \subset N$$ ($$M$$ must exist because $$N$$ is an open set and $$t$$ is in $$N$$, which makes $$t$$ an interior point of $$N$$). Since $$t$$ is a limit point of $$A$$, there exists $$q \in M \cap A$$, which implies that $$q \in N \cap A$$, as desired. QED.

Theorem. A set $$A$$ is closed if and only if $$A = \overline{A}$$.

Proof. Let $$A'$$ be a set containing all limit points of $$A$$.

Show that $$A$$ is closed only if $$A = \overline{A}$$: Since $$A$$ is closed, we have $$A' \subset A$$. So, $$A \cup A' \subset A$$. Hence, $$\overline{A} \subset A$$. Since, clearly, $$E \subset \overline{E}$$, $$E = E'$$.

Show that $$A$$ is closed if $$A = \overline{A}$$: Suppose $$A = \overline{A}$$, which implies $$A$$ contains all its limit points. Hence, $$A$$ is closed. QED.

Theorem. If $$E \subset F$$, where $$F$$ is closed, then $$\overline{E} \subset F$$.

Proof. Since $$F$$ is a closed set, $$F$$ contains all its limit point. In particular, it contains all $$E$$’s limit points. Therefore, $$F$$ contains $$\overline{E}$$. QED.

$$\overline{E}$$ is the smallest closed set containing $$E$$.

# Relationship between Open and Closed Sets

For a set $$E$$, the complement of $$E$$, denoted $$E^c$$, is $$X \setminus E = \{x : x \notin E\}$$.

Theorem. $$E$$ is open if and only if $$E^c$$ is closed.

Proof. \begin{align} E \text{ is open} & \Leftrightarrow \text{all points } p \in E \text{ has a neighborhood } N \subset E \\ & \Leftrightarrow \forall p \in E, \exists N_r(p) \subset E \\ & \Leftrightarrow \forall p \in E, \exists N_r(p) \cap E^c = \emptyset \\ & \Leftrightarrow \forall p \in E, p \text{ is not a limit point of } E^c \\ & \Leftrightarrow E^c \text{ contains all its limit points} \\ & \Leftrightarrow E^c \text{ is closed} \end{align} QED.

For example,

1. Let $$K_i = \left[-1 + \frac{1}{n}, 1 - \frac{1}{n}\right]$$ (each $$K_i$$ is closed). We have $$\bigcup_{n = 1}^\infty K_i = (-1, 1)$$, which is open and not closed.
2. Let $$K_i = \left(-\frac{1}{n}, \frac{1}{n}\right)$$ (each $$K_i$$ is open). We have $$\bigcap_{n = 1}^\infty K_i = \{0\}$$, which is closed and not open.

Lemma. Let $$\{E_\alpha\}_{\alpha \in I}$$ be a collection of sets. Then $$(\bigcup_{\alpha \in I} E_\alpha)^c = \bigcap_{\alpha \in I} E_\alpha^c$$.

Proof. \begin{align} x \in LHS & \Leftrightarrow x \notin \text{any } E_\alpha \\ & \Leftrightarrow x \in E_\alpha^c \text{ for all } \alpha \\ & \Leftrightarrow x \in \bigcap_{\alpha \in I} E_\alpha^c \end{align} QED.

Theorem. Any arbitrary union of open sets is open.

Proof. Let $$\{E_\alpha\}_{\alpha \in I}$$ be a collection of open sets. Let $$E = \bigcup_{\alpha \in I} E_\alpha$$. Consider any point $$x \in E$$. Then, $$x \in E_\beta$$ for some $$\beta \in I$$. Since $$E_\beta$$ is open, there exists a neighborhood $$N$$ around $$x$$ such that $$N \subset E_\beta$$. Hence, $$N \subset E$$. Therefore, $$x$$ is an interior point of $$E$$, as desired. QED.

Theorem. Any arbitrary intersection of closed sets is closed.

Proof. Let $$\{E_\alpha\}_{\alpha \in I}$$ be a collection of closed sets. Consider any such set $$E_\alpha$$. Since $$E_\alpha$$ is closed, $$E_\alpha^c$$ is open. From the previous theorem, we have $$\bigcup_{\alpha \in I} E_\alpha^c$$ is open. Hence, $$\left(\bigcup_{\alpha \in I} E_\alpha^c\right)^c$$ is closed. Therefore, $$\bigcap_{\alpha \in I} E_\alpha$$ is closed. QED.

Theorem. Any finite intersection of open sets is open.

Proof. Let $$E_1, E_2, \dots, E_n$$ be open sets. Let $$E = \bigcap_{i = 1}^n E_i$$. Consider any point $$x \in E$$. So, $$x \in E_i$$, for all $$1 \leq i \leq n$$. For each $$E_i$$, there exists a neighborhood $$N_{r_i}(x)$$ such that $$N_{r_i} \subset E_i$$, since $$E_i$$ is an open set. Let $$r = \min\{r_1, r_2, \dots, r_n\}$$. Then $$N_r(x)$$ is contained in $$E$$, as desired. QED.

Theorem. Any finite union of closed sets is closed.

Proof. Let $$E_1^c, E_2^c, \dots, E_n^c$$ be closed sets. Then, $$E_1^c, E_2^c, \dots, E_n^c$$ are open sets. Hence, $$\bigcap_{n = 1}^n E_i^c$$ is open. So, $$\left(\bigcap_{n = 1}^n E_i^c\right)^c$$ is closed. Thus, $$\bigcup_{n = 1}^n E_i$$ is closed. QED.

# Dense Sets

A set $$E$$ is dense in a metric space $$X$$ if every point of $$X$$ is a limit point in $$E$$ or in $$E$$. Or, equivalently, $$\overline{E} = X$$. Also, equivalently, every open set of $$X$$ contains a point in $$E$$.

For example,

1. $$\mathbb{Q}$$ is dense in $$\mathbb{R}$$.
2. Can periodic functions be approximated by sums of sines and cosines (Fourier analysis)? Or, equivalently, are linear combinations of sines and cosines dense in a set of all periodic functions? More generally, is a certain subset of functions dense in a whole space of functions?